# Thread: Prove that limit doesn't exist

1. ## Prove that limit doesn't exist

Sorry for all the questions but if there is a recurrence relation such that $x_{1} =1$, and for $n \ge 1, x_{n+1} = 3x_{n}^{2}$, how do I show that $\lim_{n\to \infty}\,(x_{n})$ does not exist. It is obvious, but I do not know how to show it. I'm assuming there is a general formula for this recurrence relation but I cannot find it. Or is there another way to prove it altogether?

2. Originally Posted by Pinkk
Sorry for all the questions but if there is a recurrence relation such that $x_{1} =1$, and for $n \ge 1, x_{n+1} = 3x_{n}^{2}$, how do I show that $\lim_{n\to \infty}\,(x_{n})$ does not exist. It is obvious, but I do not know how to show it. I'm assuming there is a general formula for this recurrence relation but I cannot find it. Or is there another way to prove it altogether?
You could just solve the recurrence relation: $x_{n+1}=3x_n^2=3\left(3x_{n-1}^2\right)^2=\cdots=3^{2^{n-1}-1}x_1=3^{2^{n-1}-1}$. Or, let $\ell_n=x_n-n$. Clearly we have that $\ell_1=1-1\geqslant 0$. Now, suppose that $\ell_n\geqslant 0$. Then, $\ell_{n+1}=x_{n+1}-(n+1)=3x_n^2-(n+1)$ $\geqslant 2x_n^2-2n\geqslant 2\left(x_n^2-n^2\right)=2\left(x_n-n\right)\cdot\left(x_n+n\right)\geqslant 3\cdot0\cdot 1=0$. It follows by induction that $\ell_n\geqslant 0\implies x_n\geqslant n$.

3. Thanks. It makes sense now but I am no good at solving recurrence relations. But I do not believe I needed to do so anyway. I just figured out that by the recurrence relation, if the limit did exist, it'd have to be either 0 or 1/3. But since every term of that sequence is greater than or equal to 1, then the limit must also be greater than or equal to 1, so obviously it does not exist.